Problem 60 - Precalculus with Calculus Previews

Question

Assume that \(2^{t}=a\) and \(6^{t}=b\). Use the laws of exponents given in this section to express the value of the given expression in terms of \(a\) and \(b\). $$ 3^{t} $$

Step-by-Step Solution

Answer

\(3^t = \frac{b}{a}\).

Step 1: Understanding the Problem

We need to express the term \(3^t\) using \(2^t = a\) and \(6^t = b\). This requires manipulating the expression using the laws of exponents.

Step 2: Expressing 6 in Terms of 2 and 3

Recall that 6 can be factored into 2 and 3. Specifically, \(6 = 2 \times 3\). Therefore, \(6^t = (2 \times 3)^t\).

Step 3: Applying the Exponent Rule

Using the law of exponents \((xy)^n = x^n y^n\), we can rewrite \((2 \times 3)^t\) as \(2^t \times 3^t\). Thus, \(6^t = 2^t \times 3^t\).

Step 4: Solving for \(3^t\)

Since \(6^t = b\) and \(2^t = a\), we have \(6^t = 2^t \times 3^t = a \times 3^t\). By dividing both sides of the equation by \(a\), we can express \(3^t\) as \(\frac{b}{a}\).

Step 5: Finalizing the Expression

Thus, the expression for \(3^t\) in terms of \(a\) and \(b\) is \(3^t = \frac{b}{a}\).

Key Concepts

Expression Manipulation

Expression manipulation is a fundamental skill in algebra. At its core, it involves altering or rearranging algebraic expressions to simplify them or put them in a useful form. This concept relies heavily on understanding algebraic rules and properties, such as the laws of exponents. For example, to express the term \(3^t\) using the relationships \(2^t = a\) and \(6^t = b\), one must keenly manipulate the expressions provided.

In our problem, identifying that \(6^t\) can be broken down into \(2^t \times 3^t\) using the exponent rule (\((xy)^n = x^n y^n\)) was crucial. This showed how the given term \(3^t\) was linked with \(2^t\) and \(6^t\). By expressing these connections clearly, we can substitute and simplify it into terms we need, revealing the relationship between variables \(a\) and \(b\). This exercise demonstrates how being adept at manipulating expressions allows you to solve complex problems by finding interconnections between known components.

Factoring

Factoring involves breaking down a composite number or expression into several often simpler factors that, when multiplied together, give the original number. In the context of exponents, this can refer to recognizing familiar bases. Within the context of exponential expressions, understanding how numbers like 6 can be decomposed into the factors 2 and 3 is invaluable.

In our scenario, identifying that 6 is equivalent to \(2 \times 3\) allowed us to express \(6^t\) as \((2 \times 3)^t\). This breakdown was essential for applying the exponent laws effectively. It introduced a method in which we could split the shared base 6 into smaller components that match our given bases in the problem (2 and 3).

• Recognizing this factorization simplifies the work involved in expression manipulations.
• Provides clarity when attempting to relate variables such as \(a\) and \(b\).

Integrating factoring into solving algebraic expressions underlines the importance of being able to see numbers not just as wholes but as products of parts.

Variable Substitution

Variable substitution is another critical technique in algebra that simplifies the process of working with expressions. It involves replacing variables with numbers or other variables to make an equation or expression easier to work with or understand.

In this exercise, given \(2^t = a\) and \(6^t = b\), substitution played a vital role when we needed to express \(3^t\) in terms of \(a\) and \(b\). By expressing \(6^t\) as \(2^t \times 3^t\), we allowed ourselves to use substitution by recognizing \(6^t\) as \(b\) and \(2^t\) as \(a\).

• Swapping out \(6^t\) for \(b\) aided in simplifying our target expression.
• Replacing \(2^t\) with \(a\) assisted in resolving \(3^t\) into \(\frac{b}{a}\).

This technique of substitution is pivotal because it allows us to change complex variables into simpler or more familiar terms, leading to quicker and often easier solutions to algebraic problems, as showcased by how we arrived at \(3^t = \frac{b}{a}\).